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andrew-mc [135]
3 years ago
10

What is a short water cycle that takes a short time to complete

Physics
2 answers:
inna [77]3 years ago
7 0
 it will be boiling water be3cause you can warm itg up 
Semenov [28]3 years ago
6 0
This could be like boiling water. it starts in liquid form. Then you boil it, and some evaporates. Then it gathers over time on the pot lid, and in some time you'll have a mini rain shower as it condenses and falls back into the pot.
You might be interested in
A man of mass 50kg ascends a flight of stairs 5m high in 5 seconds. If acceleration due to gravity is 10ms-2, the power expended
Setler79 [48]

The power expended is 500 W

Explanation:

First of all, we start by calculating the work done by the man in order to ascend: this is equal to the gravitational potential energy gained by the man, which is

W=mg\Delta h

where

m = 50 kg is the mass of the man

g=10 m/s^2 is the acceleration of gravity

\Delta h = 5 m is the change in height

Substituting,

W=(50)(10)(5)=2500 J

Now we can calculate the power expended, which is given by

P=\frac{W}{t}

where

W = 2500 J is the work done

t = 5 s is the time elapsed

Substituting, we find

P=\frac{2500}{5}=500 W

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

7 0
3 years ago
A solid cylinder of mass M = 45 kg, radius R = 0.44 m and uniform density is pivoted on a frictionless axle coaxial with its sym
user100 [1]

Answer:

w_f = 1.0345 rad/s

Explanation:

Given:

- The mass of the solid cylinder M = 45 kg

- Radius of the cylinder R = 0.44 m

- The mass of the particle m = 3.6 kg

- The initial speed of cylinder w_i = 0 rad/s

- The initial speed of particle V_pi = 3.3 m/s

- Mass moment of inertia of cylinder I_c = 0.5*M*R^2

- Mass moment of inertia of a particle around an axis I_p = mR^2

Find:

- What is the magnitude of its angular velocity after the collision?

Solution:

- Consider the mass and the cylinder as a system. We will apply the conservation of angular momentum on the system.

                                     L_i = L_f

- Initially, the particle is at edge at a distance R from center of cylinder axis with a velocity V_pi = 3.3 m/s contributing to the initial angular momentum of the system by:

                                    L_(p,i) = m*V_pi*R

                                    L_(p,i) = 3.6*3.3*0.44

                                    L_(p,i) = 5.2272 kgm^2 /s

- While the cylinder was initially stationary w_i = 0:

                                    L_(c,i) = I*w_i

                                    L_(c,i) = 0.5*M*R^2*0

                                    L_(c,i) = 0 kgm^2 /s

The initial momentum of the system is L_i:

                                    L_i = L_(p,i) + L_(c,i)

                                    L_i = 5.2272 + 0

                                    L_i = 5.2272 kg-m^2/s

- After, the particle attaches itself to the cylinder, the mass and its distribution around the axis has been disturbed - requires an equivalent Inertia for the entire one body I_equivalent. The final angular momentum of the particle is as follows:

                                   L_(p,f) = I_p*w_f

- Similarly, for the cylinder:

                                   L_(c,f) = I_c*w_f

- Note, the final angular velocity w_f are same for both particle and cylinder. Every particle on a singular incompressible (rigid) body rotates at the same angular velocity around a fixed axis.

                                  L_f = L_(p,f) + L_(c,f)

                                  L_f = I_p*w_f + I_c*w_f

                                  L_f = w_f*(I_p + I_c)

-Where, I_p + I_c is the new inertia for the entire body = I_equivalent that we discussed above. This could have been determined by the superposition principle as long as the axis of rotations are same for individual bodies or parallel axis theorem would have been applied for dissimilar axes.

                                  L_i = L_f

                                  5.2272 = w_f*(I_p + I_c)

                                  w_f =  5.2272/ R^2*(m + 0.5M)

Plug in values:

                                  w_f =  5.2272/ 0.44^2*(3.6 + 0.5*45)

                                  w_f =  5.2272/ 5.05296

                                  w_f = 1.0345 rad/s

5 0
2 years ago
Jenny was applying her makeup when she drove into the student parking lot last Friday morning . Unaware that Cheryl was stopped
Akimi4 [234]

Answer: F = 102141N

Explanation: <em><u>Newton's 2nd Law</u></em> states that a force can change the motion of a body. The relation is given by

F = m.a

whose units are:

[F] = N

[m] = kg

[a] = m/s²

Jenny's car, at the moment of the break, had acceleration:

a=\frac{\Delta v}{\Delta t}

a=\frac{11}{0.14}

a = 78.57 m/s²

Then, Force is

F = 1300*78.57

F = 102141 N

<u>Jenny's car experienced a force of </u><u>magnitude 102141N.</u>

6 0
2 years ago
20 points come an get it ppl! :D Pretend you are an electron in a copper wire which is being used to connect a battery to a flas
zaharov [31]
Batteries always use direct current. That eliminates A and C. The other part of the answers is just asking what direct/alternating means. Alternating currents change directions moving one way then another. Direct current only moves in one direction. Therefore, the answer is D, since the current is direct and direct currents move in one direction.
6 0
3 years ago
You leave the doctor's office after your annual checkup and recall that you weighed 688 N in her office. You then get into an el
Lapatulllka [165]

Answer:

a=0.5418\ m.s^{-2} upwards

a=1.283\ m.s^{-2} downwards

Explanation:

Given:

weight of the person, w=688\ N

So, the mass of the person:

m=\frac{w}{g}

m=\frac{688}{9.81}

m=70.132\ kg

  • Now if the apparent weight in the elevator, w_a= 726\ N

<u>Then the difference between the two weights is :</u>

\Delta w=w_a-w

\Delta w=726-688

\Delta w=38\ N is the force that acts on the body which generates the acceleration.

Now the corresponding acceleration:

a=\frac{\Delta w}{m}

a=\frac{38}{70.132}

a=0.5418\ m.s^{-2} upwards, because the normal reaction that due to the weight of the body is increased here.

  • Now if the apparent weight in the elevator, w_a= 598\ N

<u>Then the difference between the two weights is :</u>

\Delta w=w-w_a

\Delta w=688-598

\Delta w=90\ N is the force that acts on the body which generates the acceleration.

Now the corresponding acceleration:

a=\frac{\Delta w}{m}

a=\frac{90}{70.132}

a=1.283\ m.s^{-2} downwards, because the normal reaction that due to the weight of the body is decreased here.

7 0
3 years ago
Read 2 more answers
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